Integral

**dadon** · May 1st 2006, 02:07 AM

I am finding it very hard to solve this integration problem. Can some one explain the process step by step so that I will be able to do futher problems by myself.

Heres the question:

$\int\frac{1}{x\sqrt{4-9In^2(x)}}dx$

I know the solution but don't know how to get to it!

I hope this question is useful to others too.

Thanks

**TD!** · May 1st 2006, 02:49 AM

I assume you mean ln²(x) instead of "In". You should immediately see that ln(x)'s derivative, 1/x, is present too which makes the following substitution useful:

$ y = \ln x \Leftrightarrow dy = \frac{1}{x}dx $

This converts the integral to:

$ \int {\frac{1}{{\sqrt {4 - 9y^2 } }}dy} = \int {\frac{1}{{\sqrt {4\left( {1 - \frac{9}{4}y^2 } \right)} }}dy} = \int {\frac{1}{{2\sqrt {1 - \frac{9}{4}y^2 } }}dy} $

Now I created "sqrt(1-...)" in the denominator. If we can get that "..." to be a square, then the inverse sine should ring a bell! So let's stick that 9/4 within the square and adjust the dx (instead of another substitution):

$ \frac{1}{2}\int {\frac{1}{{\sqrt {1 - \left( {\frac{{3y}}{2}} \right)^2 } }}dy} = \frac{1}{2}\frac{2}{3}\int {\frac{1}{{\sqrt {1 - \left( {\frac{{3y}}{2}} \right)^2 } }}d\left( {\frac{{3y}}{2}} \right)} $

Now this is a standard integral, which yeilds arcsin of the integration variable.

$ \frac{1}{3}\arcsin \left( {\frac{{3y}}{2}} \right) + C \to \frac{1}{3}\arcsin \left( {\frac{3}{2}\ln x} \right) + C $

I hope this is clear

**dadon** · May 1st 2006, 03:11 AM

Hi TD!

You make maths seem so simple lol! Genius you got the right answer.
Well I get what you have done until the 3rd line where do you get the 2/3 from ? (the 2/3 multiplying with the 1/2)

Thanks

**TD!** · May 1st 2006, 03:15 AM

Actually, that is a "hidden substitution". I change the integration variable from y to 3y/2 which is allowed, if you correct with a factor 2/3 since d(y) = 2/3*3/2 d(y) = 2/3 d(3y/2). If you're uncomfortable with that, just use another substitution: 3y/2 = u <=> 3/2 dy = du <=> dy = 2/3 du, and there appears the 2/3.

**dadon** · May 1st 2006, 03:39 AM

Oh right! Thanks

I think I will better stick with another substitution! As it is probably easier for you as you can probably do it simply by looking at it.

Thanks for that.

Kind Regards

dadon

**TD!** · May 1st 2006, 03:40 AM

No problem, let us know if it doesn't work out for you

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